 # Mark the tick against the correct answer in the following: Question:

Mark the tick against the correct answer in the following:

Let $\mathrm{Q}^{+}$be the set of all positive rationals. Then, the operation $*$ on $\mathrm{Q}^{+}$defined by $\mathrm{a} * \mathrm{~b}=\frac{\mathrm{ab}}{2}$ for $\mathrm{all} \mathrm{a}, \mathrm{b} \in \mathrm{Q}^{+}$

is

A. commutative but not associative

B. associative but not commutative

C. neither commutative nor associative

D. both commutative and associative

Solution:

According to the question,

$\mathrm{Q}=\{$ Positive rationals $\}$

$\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a} * \mathrm{~b}=\mathrm{ab} / 2\}$

Formula

$*$ is commutative if $a^{*} b=b^{*} a$

$*$ is associative if $(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=\mathrm{a}^{*}\left(\mathrm{~b}^{*} \mathrm{c}\right)$

Check for commutative

Consider,$a * b=a b / 2$

And,$b * a=b a / 2$

Both equations are the same and will always true .

Therefore , * is commutative ……. (1)

Check for associative

Consider, $(\mathrm{a} * \mathrm{~b}) * \mathrm{c}=(\mathrm{ab} / 2) * \mathrm{c}=\frac{\frac{a b}{2} \times c}{2}=\mathrm{abc} / 4$

And,$a^{*}\left(b^{*} c\right)=a^{*}(b c / 2)=\frac{a \times \frac{b c}{2}}{2}=a b c / 4$

Both the equation are the same and therefore will always be true.

Therefore, $*$ is associative $\ldots \ldots$ (2)

Now, according to the equations (1), (2)

Correct option will be (D)